Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void LOG(T const& t) {
  std::cout << t << endl;
}
template <typename First, typename... Rest>
void LOG(First const& first, Rest const&... rest) {
  std::cout << first << " ";
  LOG(rest...);
}
void solve();
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  solve();
  return 0;
}
int g[110][110];
bool f;
int row[110];
int col[110];
int n, m;
int minn;
void _solve_(int x, int y) {
  for (int r = 1; r <= n; r++) {
    if (r != x) {
      row[r] = g[r][y];
    }
  }
  for (int c = 1; c <= m; c++) {
    if (c != y) {
      col[c] = g[x][c];
    }
  }
  if (min(m, n) == n) {
    for (int j = 1; j <= m; j++) {
      g[1][j] = g[1][j] + minn;
    }
  } else {
    for (int i = 1; i <= n; i++) {
      g[i][1] = g[i][1] + minn;
    }
  }
  bool fine = true;
  int total = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if ((row[i] + col[j]) != g[i][j]) {
        fine = false;
      }
    }
  }
  if (fine) {
    for (int i = 1; i <= n; i++) {
      total += row[i];
    }
    for (int j = 1; j <= m; j++) {
      total += col[j];
    }
    cout << total << endl;
    for (int i = 1; i <= n; i++) {
      if (row[i]) {
        for (int k = 0; k < row[i]; k++) {
          cout << "row " << i << endl;
        }
      }
    }
    for (int j = 1; j <= m; j++) {
      if (col[j]) {
        for (int k = 0; k < col[j]; k++) {
          cout << "col " << j << endl;
        }
      }
    }
  } else {
    cout << "-1" << endl;
  }
}
void solve() {
  cin >> n >> m;
  int x, y;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> g[i][j];
      if (g[i][j] == 0 && !f) {
        f = 1;
        x = i;
        y = j;
      }
    }
  }
  if (f) {
    _solve_(x, y);
  } else {
    int mn = min(m, n);
    minn = g[1][1];
    int _x = 1;
    int _y = 1;
    if (mn == n) {
      for (int j = 2; j <= m; j++) {
        if (minn > g[1][j]) {
          minn = g[1][j];
          _y = j;
        }
      }
    } else {
      for (int i = 2; i <= n; i++) {
        if (minn > g[i][1]) {
          minn = g[i][1];
          _x = i;
        }
      }
    }
    if (mn == n)
      row[1] = minn;
    else
      col[1] = minn;
    if (mn == n) {
      for (int j = 1; j <= m; j++) {
        g[1][j] = g[1][j] - minn;
      }
    } else {
      for (int i = 1; i <= n; i++) {
        g[i][1] = g[i][1] - minn;
      }
    }
    _solve_(_x, _y);
  }
}